Polynomial Functions, Zeros, Factors, and Intercepts

This tutorial presents a review and a collection of problems with detailed solutions on finding polynomial functions from their zeros, multiplicities, and graphs. Examples include polynomials with real coefficients and complex zeros.

Review

A)

Let \(p(x)\) be a polynomial function with real coefficients. If \[ p(s) = 0, \] then the following statements are equivalent:

  1. \(s\) is a zero of the polynomial function \(p(x)\).
  2. \(s\) is a solution of the equation \(p(x) = 0\).
  3. \((x - s)\) is a factor of \(p(x)\).
  4. The point \((s, 0)\) is an x-intercept of the graph of \(p(x)\).

B)

The imaginary unit is defined by \[ i = \sqrt{-1}. \] If \(p(x)\) has real coefficients and \(a + ib\) is a complex zero of \(p(x)\), then its complex conjugate \(a - ib\) is also a zero.

Example

Let \[ p(x) = x^3 - 2x^2 + 2x - 4. \]

  1. \[ p(2) = 2^3 - 2(2^2) + 2(2) - 4 = 8 - 8 + 4 - 4 = 0. \] Therefore, \(2\) is a zero of \(p(x)\).
  2. \(x = 2\) is a solution of \(p(x) = 0\).
  3. \(p(x)\) can be written in factored form as \[ p(x) = (x - 2)(x^2 + 2). \]
  4. The graph of \(p(x)\) has an x-intercept at \(x = 2\).
Graph of a cubic polynomial with x-intercept at x = 2

Problems with Solutions

Problem 1

The graph below represents a polynomial function \(p(x)\) with real coefficients. The degree of \(p(x)\) is 3, and all zeros are integers. Find \(p(x)\).

Graph of a cubic polynomial for Problem 1

Solution

The graph shows two x-intercepts at \(x = -1\) and \(x = 2\). The intercept at \(x = -1\) has multiplicity 2. Therefore, \[ p(x) = a(x + 1)^2(x - 2), \quad a \neq 0. \]

The y-intercept is \((0, -2)\), so \(p(0) = -2\): \[ a(1)^2(-2) = -2. \] Solving gives \(a = 1\).

Hence, \[ p(x) = (x + 1)^2(x - 2). \]

Problem 2

A polynomial function \(p(x)\) of degree 5 with real coefficients has the zeros \(-1\), \(2\) (with multiplicity 2), \(0\), and \(1\). If \(p(3) = -12\), find \(p(x)\).

Solution

The polynomial can be written as \[ p(x) = a\,x(x + 1)(x - 2)^2(x - 1), \quad a \neq 0. \]

Using \(p(3) = -12\): \[ a(3)(4)(1)^2(2) = -12. \] This gives \(a = -\tfrac{1}{2}\).

Therefore, \[ p(x) = -\tfrac{1}{2}x(x + 1)(x - 2)^2(x - 1). \]

Problem 3

The number \(2 + i\) is a zero of the polynomial \[ p(x) = x^4 - 2x^3 - 6x^2 + 22x - 15. \] Find all other zeros.

Solution

Since \(p(x)\) has real coefficients, the conjugate \(2 - i\) is also a zero. Then \[ p(x) = (x - (2 + i))(x - (2 - i))q(x). \]

Compute \[ (x - (2 + i))(x - (2 - i)) = x^2 - 4x + 5. \]

Dividing \(p(x)\) by \(x^2 - 4x + 5\) gives \[ q(x) = x^2 + 2x - 3. \]

Factor: \[ x^2 + 2x - 3 = (x - 1)(x + 3). \]

The zeros are \[ 2 + i, \; 2 - i, \; 1, \; -3. \]

Problem 4

The number \(-3 - i\) is a zero of the polynomial \[ p(x) = x^4 + 6x^3 + 11x^2 + 6x + 10. \] Find all other zeros.

Solution

Because the coefficients are real, the conjugate \(-3 + i\) is also a zero. Then \[ p(x) = (x + 3 + i)(x + 3 - i)q(x). \]

Compute \[ (x + 3 + i)(x + 3 - i) = x^2 + 6x + 10. \]

Dividing gives \[ q(x) = x^2 + 1. \]

Solve \[ x^2 + 1 = 0 \quad \Rightarrow \quad x = \pm i. \]

The zeros of \(p(x)\) are \[ -3 - i, \; -3 + i, \; i, \; -i. \]

More References on Polynomial Functions